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Old 2005-08-29, 19:44   #1
Xyzzy
 
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Exclamation Math rendering engine

\2^{57,885,161}-1

This is using our new \small \TeX tags... Look at the last entry here:

http://www.mersenneforum.org/misc.php?do=bbcode

Manual:

http://www.mersenneforum.org/mimetex.htm
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Old 2005-08-29, 20:06   #2
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Great!! now we can write things like e^{i\pi}+1=0 and \sum_{i=1}^{n}x_i=\int_0^{1}f.

We need a few good links to online tex guides...
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Old 2005-08-29, 20:08   #3
xilman
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Quote:
Originally Posted by Xyzzy
\2^p-1

This is using our new \small \TeX tags... Look at the last entry here:

http://www.mersenneforum.org/misc.php?do=bbcode
Excellent!

I've long wanted to be able to use \TeX formatting.


Paul
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Old 2005-08-29, 20:18   #4
Xyzzy
 
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There are several reasons why this took so long to get set up. The first reason is our old server didn't run Linux so a lot of software wasn't available. Secondly, we had very limited access to the server which prevented us from doing really cool stuff. We were also was hesitant to dive into a solution that didn't have a workable license. As it is, it pains us that the forum software itself isn't GPLed.

We have no idea how much of a load this will put on the server, but at any time we can redirect the image generation to another box running this. For now, especially if you are going to test some really crazy stuff, use the interactive editor thing.

Now that we have this set up, we realize we could have had this a long time ago, by redirecting the image generation to another box. We apologize that it took us this long to figure it out.

Who is going to post the most interesting example formula?

Have fun!
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Old 2005-08-29, 23:30   #5
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Just testing.

\displaystyle x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}

Could be bigger, don't you think?

Last fiddled with by Numbers on 2005-08-29 at 23:31
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Old 2005-08-29, 23:39   #6
Xyzzy
 
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Quote:
Originally Posted by Numbers
Could be bigger, don't you think?
\LARGE \displaystyle x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}
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Old 2005-08-30, 00:00   #7
Numbers
 
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\displaystyle2\left(\frac12+\frac23\right)=\frac53



Xyzzy, I think you're an absolute genius.
But you're going to have to tell me how to make it bigger, because even in displaystyle it still comes out pretty small.
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Old 2005-08-30, 00:03   #8
Xyzzy
 
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Quote:
Originally Posted by Numbers
But you're going to have to tell me how to make it bigger, because even in displaystyle it still comes out pretty small.
Just hit "Quote" to see the source and you will see we added a \LARGE command. The documentation covers this. We have a sneaky feeling we could spend a lot of time learning all this. It seems to be very powerful. The interactive editor we linked to earlier makes it easier to try new stuff quickly.
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Old 2005-08-30, 00:14   #9
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Quote:
Originally Posted by Numbers
\displaystyle2\left(\frac12+\frac23\right)=\frac53
Big, or small, I don't like the math.
\LARGE\displaystyle2\left(\frac12+\frac23\right)=\frac73
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Old 2005-08-30, 02:04   #10
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So there is more than one type of Latex?

\displaystyle{f(x)=f(a)+f'(a)x+\frac{1}{2}f''(a)x^2+\frac{1}{6}f'''(a)x^3+...}

(I guess that answers my question. What I just typed would surely have worked on a certain other forum).

Last fiddled with by jinydu on 2005-08-30 at 02:05
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Old 2005-08-30, 04:32   #11
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Often I use congruences, so I'll just pick a theorem at random...

Theorem: Let p be a prime such that p^2 divides M_q=2^q-1 for some prime q. Then 2^{\frac{p-1}{2}} \equiv 1 \pmod {p^2} and so p is a Wieferich prime.

Proof: We know that if p divides M_q=2^q-1, then p=2kq+1 for some k \in \mathbb{N}. Then q=\frac{p-1}{2k}, and if p^2 divides M_q=2^q-1 then 2^q \equiv 2^{\frac{p-1}{2k}} \equiv 1 \pmod {p^2}.
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